Computer Graphics 5 - View transformation andclipping

Tom Thorne

Slides courtesy of Taku Komurawww.inf.ed.ac.uk/teaching/courses/cg

Overview

View transformation

I Homogeneous coordinates recapI Parallel projectionI Persepctive projectionI Canonical view volume

Clipping

I Line and polygon clipping

Outline of procedure

1. Transform from world to camera coordinates2. Project into view volume or screen coordinates

3. Clip geometry that falls outside of the view volume

vg

g

c

Mg<-c

View projection

I Homogeneous transformationI Parallel projectionI Perspective projectionI Canonical view volume

Homogeneous transformations

v = Mproj

McΩg

MgΩl

vl

vg

g

c

Mg<-c vgv l

v l

Projection Global to camera Local to global

Homogeneous coordinates

I Allows us to represent translation, rotation and scaling bymatrix multiplication

I Coordinates (x , y , z , w) and (cx , cy , cz , cw) represent the samepoint

I Return to Cartesian coordinates using(x Õ, y

Õ, z

Õ) = (x/w , y/w , z/w)

Camera coordinate system

I As used by OpenGLI Camera is facing in the negative z direction

z

y

x

y

x

(1,1)(-1,1)

(-1,-1) (1,-1)

Parallel (orthographic) projection

I Specified by a direction of projectionI No change in size of objects

Parallel projection

Project coordinates onto plane at z = 0x

Õ = x , y

Õ = y , z

Õ = 0

M

orth

=

Q

ccca

1 0 0 00 1 0 00 0 0 00 0 0 1

R

dddb

Q

ccca

1 0 0 00 1 0 00 0 0 00 0 0 1

R

dddb

Q

ccca

x

y

z

1

R

dddb =

Q

ccca

x

y

01

R

dddb

Perspective projection

I Far away objects appear smallerI Specified by a center of projection and focal distance

Perspective projection

y

x

+z -z

d

(x,y,z)

(x ,y ,-d)p p

Perspective projection - simple case

From similar triangles:

I x

p

d

= x

≠z

, y

p

d

= y

≠z

Ix

p

= dx

≠z

= x

≠z/d

, y

p

= dy

≠z

= y

≠z/d

y

+z -z

d

(x,y,z)

(x ,y ,-d)p p

yp

y

Perspective projection

Q

ccca

x

Õ

y

Õ

≠d

1

R

dddb =

Q

ccca

≠d

x

z

≠d

y

z

≠d

1

R

dddb =

Q

ccca

x

y

z

≠z/d

R

dddb

Represented as a matrix multiplication:

M

proj

=

Q

ccca

1 0 0 00 1 0 00 0 1 00 0 ≠ 1

d

0

R

dddb

Perspective projection

V

p

= M

proj

V

c

=

Q

ccca

1 0 0 00 1 0 00 0 1 00 0 ≠ 1

d

0

R

dddb

Q

ccca

x

y

z

1

R

dddb =

Q

ccca

x

y

z

≠ z

d

R

dddb =

Q

ca

x

≠z/d

y

≠z/d

≠d

R

db

Alternative formulationCenter of projection at z = d , projection plane at z = 0

I x

p

d

= x

≠z+d

, y

p

d

= y

≠z+d

Ix

p

= dx

≠z+d

= x

(≠z/d)+1 , y

p

= dy

≠z

= y

(≠z/d)+1

y

+z -z

d

(x,y,z)

(x ,y ,0)p p

yp

y

This allows for d æ Œ

Alternative formulationy

+z -z

d

(x,y,z)

(x ,y ,0)p p

yp

y

Ix

p

= dx

≠z+d

= x

(≠z/d)+1 , y

p

= dy

≠z

= y

(≠z/d)+1

M

proj

=

Q

ccca

1 0 0 00 1 0 00 0 0 00 0 ≠1

d

1

R

dddb

Exercise

Where will the two points be projected?

(-1,-1,-1)

(-2,2,-2)

d=1

M

proj

=

Q

ccca

1 0 0 00 1 0 00 0 0 00 0 ≠ 1

d

1

R

dddb

Problems

I After projection depth information is lost (this is needed forhidden surface removal)

I Objects behind the camera are projected infront of the camera!

3D view volumeDefine a volume in which objects are visible

I For parallel projection, a boxI For perspective projection, a frustumI Surfaces outside this volume are clipped or removed and not

drawn

Canonical view volume

It can be computationally expensive to check if a point is inside afrustum

I Instead transform the frustum into a normalised canonical viewvolume

I Uses the same ideas a perspective projectionI Makes clipping and hidden surface calculation much easier

Transforming the view frustumThe frustum is defined by a set of parameters, l , r , b, t, n, f :

l Left x coordinate ofnear plane

r Right x coordinateof near plane

b Bottom ycoordinate of nearplane

t Top y coordinate ofnear plane

n Minus z coordinateof near plane

f Minus z coordinateof far plane

With 0 < n < f .

z=-n

z=-f

(l,b,-n) (r,t,-n)

Transforming the view frustum

We can transform the perspective canonical view volume to theparallel view volume using:

If z œ [≠n, ≠f ] and 0 < n < f then

M

can

=

Q

ccca

2n

r≠l

0 r+l

r≠l

00 2n

t≠b

t+b

t≠b

00 0 ≠ f +n

f ≠n

≠ 2fn

f ≠n

0 0 ≠1 0

R

dddb

Final steps

I After transformation to canonical view volume, divide by w toproduce 3D Cartesian coordinates

I Perform clipping in the canonical view volume:

≠1 Æ x Æ 1,≠1 Æ y Æ 1,≠1 Æ z Æ 1

Example

What does ABC look like after projection?

M

can

=Q

ccca

2n

r≠l

0 r+l

r≠l

00 2n

t≠b

t+b

t≠b

00 0 ≠ f +n

f ≠n

≠ 2fn

f ≠n

0 0 ≠1 0

R

dddb

ExampleM

can

=Q

ccca

2n

r≠l

0 r+l

r≠l

00 2n

t≠b

t+b

t≠b

00 0 ≠ f +n

f ≠n

≠ 2fn

f ≠n

0 0 ≠1 0

R

dddb =

Q

ccca

1 0 0 00 1 0 00 0 ≠2 ≠30 0 ≠1 0

R

dddb

n = 1,f = 3,r = 1,l = ≠1,t =1,b = ≠1

Q

ccca

1 0 0 00 1 0 00 0 ≠2 ≠30 0 ≠1 0

R

dddb

Q

ccca

30

≠21

R

dddb =

Q

ccca

3012

R

dddb =

Q

ccca

320121

R

dddb

Projection summaryI Parallel and perspective projectionI Projection matrices transform points to 2D coordinates on the

screenI Canonical view volumes can be used for clipping

z=-n

z=-f

(l,b,-n) (r,t,-n)

Overview

View transformation

I Homogeneous coordinates recapI Parallel projectionI Persepctive projectionI Canonical view volume

Clipping

I Line and polygon clipping

Projection of polygons and lines

I Lines in 3D become lines in 2DI Polygons in 3D become polygons in 2D

Clipping

They may intersect the canonical view volume, then we need toperform clipping:

I Clipping lines (Cohen-Sutherland algorithm)I Clipping polygons (Sutherland-Hodgman algorithm)

Cohen-Sutherland algorithm

Input Screen and 2D line segmentOutput Clipped 2D line segment

Cohen-Sutherland algorithm

0001

101010001001

0010

011001000101

0000

Outcodes:

I Space is split into nine regionsI The center corresponds to the area visible on the screenI Each region is encoded by four bits

Cohen-Sutherland algorithm

0001

101010001001

0010

011001000101

0000

4th bit

1st bit

2nd bit

3rd bit

Outcodes:

1st bit Above top of screen (y > y

max

)2nd bit Below bottom of screen (y < y

min

)3rd bit Right of right edge of screen (x > x

max

)4th bit Left of left edge of screen (x < x

min

)

Cohen-Sutherland algorithm

Data: Line from coordinate A to BResult: Clipped linewhile true do

Calculate outcodes of endpoints;if trivial accept or trivial reject then

return accepted line or reject;else

Find which endpoint is outside clipping rectangle;Clip outside endpoint to edge of first non-zero outcode bitand update endpoint coordinates;

end

end

Bitwise AND and OR

0 OR 0 = 00 OR 1 = 11 OR 0 = 11 OR 1 = 1

0 AND 0 = 00 AND 1 = 01 AND 0 = 01 AND 1 = 1

Trivial acceptAll line endpoints are inside the screen

I Apply an OR to outcodes of the line endpoints, test for equalto 0

Cohen-Sutherland algorithm

Data: Line from coordinate A to BResult: Clipped linewhile true do

Calculate outcodes of endpoints;if trivial accept or trivial reject then

return accepted line or reject;else

Find which endpoint is outside clipping rectangle;Clip outside endpoint to edge of first non-zero outcode bitand update endpoint coordinates;

end

end

Trivial rejectBoth line endpoints are outside of the screen on the same side

I Apply AND operation to the two endpoints and reject if not 0

Cohen-Sutherland algorithm

0001

101010001001

0010

011001000101

0000

0001

0000

1000

0000

Logical AND between codes for two endpoints. If non-zero, trivialrejection.

Cohen-Sutherland algorithm

0001

101010001001

0010

011001000101

0000

Logical AND between codes for two endpoints. If non-zero, trivialrejection.

Cohen-Sutherland algorithm

Data: Line from coordinate A to BResult: Clipped linewhile true do

Calculate outcodes of endpoints;if trivial accept or trivial reject then

return accepted line or reject;else

Find which endpoint is outside clipping rectangle;Clip outside endpoint to edge of first non-zero outcode bitand update endpoint coordinates;

end

end

Line intersectionI Need to clip a line to the edges of the rectangleI Find an endpoint which is outside the rectangleI Select edge to clip based on the outcode, split the line and

feedback into algorithm

A

B

A'A''

Cohen-Sutherland algorithm

Extension to 3D

I Clip lines to near and far planes of view volumeI Needs more bits for the outcode

Polygon clipping

Sutherland-Hodgman algorithm

Input 2D polygonOutput List of clipped vertices

Traverse the polygon and clip at each edge of the screen

Sutherland-Hodgman algorithm

For each edge of the clipping rectangle:

I For each polygon edge between v

i

and v

i+1

Edge of screenOutsideInside

Case 1

Output both vertices

Edge of screenOutsideInside

Case 2

Output intersection with screen

Edge of screenOutsideInside

Case 3

No output

Edge of screenOutsideInside

Case 4

Output intersection with screenand second vertex of edge

Example

Summary

Projection

I Parallel and perspectiveI Canonical view volumes

Clipping

I Cohen-Sutherland algorithmI Sutherland-Hodgman algorithm

References

View transformation

I Shirley, Chapter 7I Foley, Chapter 6

Clipping

I Foley Chapter 3.12, 3.14I http://www.cc.gatech.edu/grads/h/Hao-wei.Hsieh/

Haowei.Hsieh/mm.html